Question 150308
square root (x+6)+square root (X-2) = 4
(x+6)2+(x-2)2
X2+6x+6x+36+x2-2x-2x+4=42
2x2+8x+40=16
2x2+8x+24=0
(2x+…)(x+…) 
I cannot find 2 numbers that equal 24 but also 8 in the middle... did I approach this exercise wrong? The task was just to solve it...
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{{{sqrt(x+6)+sqrt(X-2) = 4}}}
Square both sides
{{{(x+6) + 2*sqrt(x+6)*sqrt(x-2) +(x-2) = 16}}}
Collect terms
{{{2x + 4 + 2*sqrt(x+6)*sqrt(x-2) = 16}}}
Isolate radicals
{{{2*sqrt(x+6)*sqrt(x-2) = 12 - 2x}}}
Divide by 2
{{{sqrt(x+6)*sqrt(x-2) = 6 - x}}}
Square both sides again
{{{(x+6)*(x-2) = x^2 - 12x + 36}}}
expand
{{{x^2 + 4x -12 = x^2 -12x + 36}}}
Collect terms
16x = 48
x = 3
---- Sub in 3 for x to check
{{{sqrt(x+6)+sqrt(X-2) = 4}}}
{{{sqrt(3+6)+sqrt(3-2) = 4}}}
3 + 1 = 4
Looks good.
You were squaring the binomials inside the radical, shoulda squared just the radical in the first step.